What happens to Protons and Electrons when a Neutron star forms? At some point gravity overcomes the Pauli Exclusion Principle ( I assume) and they are all forced together. What happens in the process?
[Physics] What happens to Protons and Electrons when a Neutron star forms
astrophysicsneutron-starsnuclear-physics
Related Solutions
How does the Pauli Exclusion Principle actually create a force?
The Pauli exclusion principle doesn't really say that two fermions can't be in the same place. It's both stronger and weaker than that. It says that they can't be in the same state, i.e., if they're standing waves, two of them can't have the same standing wave pattern. But for bulk matter, for our purposes, it becomes a decent approximation to treat the exclusion principle as saying that if $n$ particles are confined to a volume $V$, they must each be confined to a space of about $V/n$. Since volume goes like length cubed, this means that their wavelengths must be $\lesssim (V/n)^{1/3}$. As $V$ shrinks, this maximum wavelength shrinks as well, and the de Broglie relation then tells us that the momentum goes up. The increased momentum shows up as a pressure, just as it would if you increased the momenta of all the molecules in a sample of air. A degenerate body like a neutron star or white dwarf is in a state where this pressure is in equilibrium with gravity.
[Physics] Why does this quiz question say that protons and electrons do not combine to form neutrons
You're asking about two distinct phenomena. The difference between them is subtle, and I think there is some context missing from the second question that you quote, which makes things more confusing than they need to be.
When the neutron star forms, most of the protons and electrons combine together to form neutrons
This is mostly correct. The process is known as "electron capture," and the full reaction is
$$\rm p + e^- \to n + \nu_e$$
The other particle in the final state (represented by a nu) is a neutrino. The neutrino is an uncharged, very low-mass electron-like particle, in the same way that neutrons and protons are different charge states of the same sort of particle. So far as we know, in physics, the number of electron-like "leptons" and the number of proton-like "baryons" isn't changed in any physical process. The neutrinos play an important role in the dynamics of the stellar collapses where neutron stars are formed, but in some authors who write very elementary explanations of neutron stars will leave the neutrinos out of their descriptions. There are advantages and disadvantages to this approach; your confusion here is one of the disadvantages.
A neutron is formed by an electron and a proton combining together, therefore it is neutral: true or false? Answer: false
This is a fundamentally flawed true-false question, because it makes several statements at the same time, some of which are correct. The question I was expecting to find here, based on the title of your question, was more like
The neutron is an electron and a proton that are "stuck together" somehow. (Answer: false)
We have another name for an electron and a proton that are semi-permanently "stuck together," and the dynamics of that system are very different from the dynamics of the neutron.
When you "combine together" macroscopic objects in ordinary life, the things that you combined are still somehow present in the combination. But in particle physics, the situation is different. The electron-capture process that we're talking about here fundamentally changes both the baryon and the lepton parts of the system. To the extent that a neutron behaves like a composite particle, it behaves as if it is made out of quarks.
Best Answer
It is the Pauli Exclusion Principle that actually allows the formation of a "neutron" star.
In an "ordinary" gas of protons and electrons, nothing would happen - we call that ionized hydrogen! However, when you squeeze, lots of interesting things happen. The first is that the electrons become "degenerate". The Pauli exclusion principle forbids more than two electrons (one spin up the other spin down) from occupying the same momentum eigenstate (particles in a box occupy quantised momentum states).
In that case what happens is that the electrons "fill up" the low momentum/low energy states and are then forced to fill increasingly higher momentum/energy states. The electrons with large momentum consequently exert a degeneracy pressure, and it is this pressure that supports white dwarf stars.
If the density is increased even further - the energies of degenerate electrons at the top of the momentum/energy distribution get so large that they are capable of interacting with protons (via the weak nuclear force) in a process called inverse beta decay (sometimes referred to as electron capture when the proton is part of a nucleus) to produce a neutron and a neutrino. $$p + e \rightarrow n + \nu_e$$ Ordinarily, this endothermic process does not occur, or if it does, the free neutron decays back into a proton and electron. However at the densities in a neutron star, not only can the degenerate electrons have sufficient energy to instigate this reaction, their degeneracy also blocks neutrons from decaying back into electrons and protons. The same is also true of the protons (also fermions), which are also degenerate at neutron star densities.
The net result is an equilibrium between inverse beta decay and beta decay. If too many neutrons are produced, the drop in electron/proton densities leaves holes at the top of their respective energy distributions that can be filled by decaying neutrons. However if too many neutrons decay, the electrons and protons at the tops of their respective energy distributions have sufficient energies to create new neutrons.
Mathematically, this equilibrium is expressed as $$E_{F,p} + E_{F,e} = E_{F,n},$$ where these are the "Fermi energies" of the degenerate protons, electrons and neutrons respectively, and we have the additional constraint that the Fermi momenta of the electrons and protons are identical (since their number densities would be the same).
At neutron star densities (a few $\times 10^{17}$ kg/m$^{3}$) the ratio of neutrons to protons is of order 100. (The number of protons equals the number of electrons).
This calculation assumes ideal (non-interacting) fermion gases. At even higher densities (cores of neutron stars) the strong interaction between nucleons in the asymmetric nuclear matter alters the equilibrium above and causes the n/p ratio to decrease to closer to 10.